珂朵莉树

珂朵莉树是一种对针对随机数据“投机取巧”的暴力， 核心是因为大量的区间赋值操作会导致产生许多满足元素全部相等且长度较大的区间。所以采用存储区间的方式。
随机数据复杂度O(logn) ，然而可能被卡导致退化成链表O(n)。
(模板题)[https://www.luogu.com.cn/problem/CF896C]
(代码转载自洛谷的题解)[https://www.luogu.com.cn/article/v75h91bh]，并增加了注释。

#include <iostream>
#include <set>
#include <algorithm>
#include <vector>
#include <cstdio>

using namespace std;

typedef long long ll;
const ll MOD = 1000000007;
const ll MAXN = 100005;

struct Node {
    ll l, r;//l和r表示这一段的起点和终点
    mutable ll v;//v表示这一段上所有元素相同的值是多少

    Node(ll l, ll r = 0, ll v = 0) : l(l), r(r), v(v) {}

    bool operator<(const Node &a) const {
        return l < a.l;//规定按照每段的左端点排序
    }
};

ll n, m, seed, vmax, a[MAXN];
set<Node> s;

//以pos去做切割，找到一个包含pos的区间，把它分成[l,pos-1],[pos,r]两半
auto split(int pos) {
    auto it = s.lower_bound(Node(pos));
    if (it != s.end() && it->l == pos) {
        return it;
    }
    it--;
    if (it->r < pos) return s.end();
    ll l = it->l;
    ll r = it->r;
    ll v = it->v;
    s.erase(it);
    s.insert(Node(l, pos - 1, v));
    //insert函数返回pair，其中的first是新插入结点的迭代器
    return s.insert(Node(pos, r, v)).first;
}


void add(ll l, ll r, ll x) {
    auto itr = split(r + 1), itl = split(l);
    for (auto it = itl; it != itr; ++it) {
        it->v += x;
    }
}

void assign(ll l, ll r, ll x) {
    auto itr = split(r + 1), itl = split(l);
    s.erase(itl, itr);
    s.insert(Node(l, r, x));
}

struct Rank {
    ll num, cnt;

    bool operator<(const Rank &a) const {
        return num < a.num;
    }

    Rank(ll num, ll cnt) : num(num), cnt(cnt) {}
};

ll rnk(ll l, ll r, ll x) {
    auto itr = split(r + 1), itl = split(l);
    vector<Rank> v;
    for (auto i = itl; i != itr; ++i) {
        v.push_back(Rank(i->v, i->r - i->l + 1));
    }
    sort(v.begin(), v.end());
    int i;
    for (i = 0; i < v.size(); ++i) {
        if (v[i].cnt < x) {
            x -= v[i].cnt;
        } else {
            break;
        }
    }
    return v[i].num;
}

ll ksm(ll x, ll y, ll p) {
    ll r = 1;
    ll base = x % p;
    while (y) {
        if (y & 1) {
            r = r * base % p;
        }
        base = base * base % p;
        y >>= 1;
    }
    return r;
}

ll calP(ll l, ll r, ll x, ll y) {
    auto itr = split(r + 1), itl = split(l);
    ll ans = 0;
    for (auto i = itl; i != itr; ++i) {
        ans = (ans + ksm(i->v, x, y) * (i->r - i->l + 1) % y) % y;
    }
    return ans;
}

ll rnd() {
    ll ret = seed;
    seed = (seed * 7 + 13) % MOD;
    return ret;
}

int main() {
    cin >> n >> m >> seed >> vmax;
    for (int i = 1; i <= n; ++i) {
        a[i] = (rnd() % vmax) + 1;
        s.insert(Node(i, i, a[i]));
    }
    for (int i = 1; i <= m; ++i) {
        ll op, l, r, x, y;
        op = (rnd() % 4) + 1;
        l = (rnd() % n) + 1;
        r = (rnd() % n) + 1;
        if (l > r) swap(l, r);
        if (op == 3) {
            x = (rnd() % (r - l + 1)) + 1;
        } else {
            x = (rnd() % vmax) + 1;
        }
        if (op == 4) {
            y = (rnd() % vmax) + 1;
        }
        if (op == 1) {
            add(l, r, x);
        } else if (op == 2) {
            assign(l, r, x);
        } else if (op == 3) {
            cout << rnk(l, r, x) << endl;
        } else {
            cout << calP(l, r, x, y) << endl;
        }
    }
    return 0;
}